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Pyrolusite, an ore of manganese, contains manganese in the form of MnO₂ - VCE - SSCE Chemistry - Question 7 - 2003 - Paper 1

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Pyrolusite, an ore of manganese, contains manganese in the form of MnO₂. A sample of pyrolusite from a newly discovered deposit is analysed to determine the degree o... show full transcript

Worked Solution & Example Answer:Pyrolusite, an ore of manganese, contains manganese in the form of MnO₂ - VCE - SSCE Chemistry - Question 7 - 2003 - Paper 1

Step 1

Calculate the amount in mole of oxalic acid remaining in the original 100 mL solution after the pyrolusite had been reacted with the oxalic acid.

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Answer

First, calculate the initial moles of oxalic acid present in the 100 mL solution:

n(H2C2O4)=0.150imes0.100=0.0150 moln(H₂C₂O₄) = 0.150 imes 0.100 = 0.0150 \text{ mol}

Next, calculate the amount of oxalic acid that reacted with I₃⁻. First, find the moles of I₃⁻:

n(I3−)=0.050×0.022=0.00110 moln(I₃⁻) = 0.050 \times 0.022 = 0.00110 \text{ mol}

From the stoichiometry of the reaction, 1 mole of H₂C₂O₄ reacts with 3 moles of I₃⁻:

n(H2C2O4) used=0.001103=0.000367 moln(H₂C₂O₄)\text{ used} = \frac{0.00110}{3} = 0.000367 \text{ mol}

Now, subtract the moles of H₂C₂O₄ used from the initial amount:

n(H2C2O4) remaining=0.0150−0.000367=0.014633 moln(H₂C₂O₄)\text{ remaining} = 0.0150 - 0.000367 = 0.014633 \text{ mol}

Step 2

Calculate the amount in mole of oxalic acid used to reduce the MnO₂ in 1.25 g of pyrolusite.

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Answer

The moles of oxalic acid originally present were calculated as:

n(H2C2O4) used=0.0150−n(H2C2O4) remaining=0.0150−0.014633=0.000367 moln(H₂C₂O₄)\text{ used} = 0.0150 - n(H₂C₂O₄)\text{ remaining} = 0.0150 - 0.014633 = 0.000367 \text{ mol}

Thus, the amount of oxalic acid used to reduce MnO₂ is approximately 0.000367 moles.

Step 3

Calculate the amount in mole of MnO₂ present in the original 1.25 g of pyrolusite and hence the percentage of MnO₂ by mass present in the pyrolusite.

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Answer

From the reaction, 1 mole of MnO₂ reacts with 1 mole of H₂C₂O₄, so the amount of MnO₂ can be calculated as:

n(MnO2)=n(H2C2O4) used=0.000367 moln(MnO₂) = n(H₂C₂O₄)\text{ used} = 0.000367 \text{ mol}

Next, calculate the molar mass of MnO₂:

M(MnO2)=54.94+(16.00×2)=86.94 g/molM(MnO₂) = 54.94 + (16.00 \times 2) = 86.94 \text{ g/mol}

Now, calculate the mass of MnO₂:

m(MnO2)=0.000367 mol×86.94 g/mol≈0.03195 gm(MnO₂) = 0.000367 \text{ mol} \times 86.94 \text{ g/mol} \approx 0.03195 \text{ g}

Finally, calculate the percentage of MnO₂ by mass in the 1.25 g sample:

Percentage=(0.03195 g1.25 g)×100≈2.56%\text{Percentage} = \left( \frac{0.03195 \text{ g}}{1.25 \text{ g}} \right) \times 100 \approx 2.56\%(rounded to 2 decimal places).

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